Loop and getchar problems

[Samuel.Coddler]

Hi,

I'm a newbie programmer. I can't get work the following code.

/*
Objective:

Use a loop to print out all of the input characters until a newline is
found.

*/
#include <stdio.h>

int main (){
/* Declare variable, Assign a initial value x. */
char a = 'x';

while (a != '\n') {
/* Get a user input */
a = getchar();

if (a == '\n'){
/* If the input is a new line, say good bye to user */
printf("\nThat's it for today.\n");

} else {
/* Prints a user input */
printf("%c",a);
}

}

return 1;
}

[Tak]

On 6ÔÂ1ÈÕ, ÏÂÎç5ʱ25·Ö, [e-mail] wrote:

Hi,

I'm a newbie programmer. I can't get work the following code.

/*
Objective:

Use a loop to print out all of the input characters until a newline is
found.

*/

#include <stdio.h>

int main (){
/* Declare variable, Assign a initial value x. */
char a = 'x';

while (a != '\n') {

/* Get a user input */
a = getchar();

if (a == '\n'){
/* If the input is a new line, say good bye to user */
printf("\nThat's it for today.\n");

} else {
/* Prints a user input */
printf("%c",a);
}

}

return 1;

}

It's my program:
#include <stdio.h>

int main (){
char a;

while ((a = getchar()) != '\n') {
printf("%c",a);
}
printf("\nThat's it for today.\n");

return 1;
}

right??????

[mark_bluemel]

On 1 Jun, 10:25, [e-mail] wrote:

Hi,

I'm a newbie programmer. I can't get work the following code.

"I can't get work" doesn't tell me much about your problem.

Generally if there's a problem you need help with, it's helpful to
tell us
* What did you expect to happen?
* What did happen?

Your code works for me, and behaves exactly as I expect, but that's
not necessarily what you expected, so you probably need to tell us
what you expected.

I'll give you a hint - standard input is probably line-buffered...

>
/*
Objective:

Use a loop to print out all of the input characters until a newline is
found.

*/

#include <stdio.h>

int main (){
/* Declare variable, Assign a initial value x. */
char a = 'x';

while (a != '\n') {

/* Get a user input */
a = getchar();

if (a == '\n'){
/* If the input is a new line, say good bye to user */
printf("\nThat's it for today.\n");

} else {
/* Prints a user input */
printf("%c",a);
}

}

return 1;

}

[Chris Dollin]

Sa************@gmail.com wrote:

Minor notes (since Mark has started the major ones):

#include <stdio.h>

int main (){
/* Declare variable, Assign a initial value x. */

Pointless comment.

char a = 'x';

while (a != '\n') {
/* Get a user input */

Ditto.

a = getchar();

`getchar` returns `int`. So `a` should be `int`. (This is because
the result of `getchar` has to allow for every character /and/
EOF.)

(fx:snip)

return 1;

Not portable. 0, EXIT_SUCCESS, EXIT_FAILURE (those from stdlib)
are portable.

}

--
"- born in the lab under strict supervision -", - Magenta, /Genetesis/

Hewlett-Packard Limited Cain Road, Bracknell, registered no:
registered office: Berks RG12 1HN 690597 England

[Samuel.Coddler]

Hi, Tak,

Your code is much clearner. But the result on my end is the same.

I wanna make the program so that the program prints all the inputs
until a user input a new like (generally by pressing return).

So, it should open for a new input.
That string "That's it for today" should be displayed only when a user
type a newline.

[Richard Heathfield]

Tak said:

It's my program:
#include <stdio.h>

int main (){
char a;

while ((a = getchar()) != '\n') {
printf("%c",a);
}
printf("\nThat's it for today.\n");

return 1;
}

right??????

Wrong. Well, you're right that it's your program, but you're wrong that
it's right. It's wrong.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

[mark_bluemel]

On 1 Jun, 10:53, [e-mail] wrote:

On 1 Jun, 10:25, [e-mail] wrote:

Hi,

I'm a newbie programmer. I can't get work the following code.

"I can't get work" doesn't tell me much about your problem.

Generally if there's a problem you need help with, it's helpful to
tell us
* What did you expect to happen?
* What did happen?

Your code works for me, and behaves exactly as I expect, but that's
not necessarily what you expected, so you probably need to tell us
what you expected.

[Snip]

I'd better qualify my comments by saying that I didn't bother with the
nitpicking about whether you should declare main as "int main(void)"
or that a should be an int.

[Richard Heathfield]

ma**********@pobox.com said:

On 1 Jun, 10:25, [e-mail] wrote:

>Hi,

I'm a newbie programmer. I can't get work the following code.

"I can't get work" doesn't tell me much about your problem.

Actually, I think it tells you all you need to know.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

[Richard Heathfield]

Sa************@gmail.com said:

Hi,

I'm a newbie programmer. I can't get work the following code.

Here's your program again, properly indented for readability:

#include <stdio.h>

int main()
{
/* Declare variable, Assign a initial value x. */
char a = 'x';

while(a != '\n')
{
/* Get a user input */
a = getchar();

if(a == '\n')
{
/* If the input is a new line, say good bye to user */
printf("\nThat's it for today.\n");
}
else
{
/* Prints a user input */
printf("%c", a);
}
}

return 1;
}

Problems:

1) a has a lousy name!
2) a has the wrong type - it should be int.

Whilst I wouldn't want to suggest that your program is otherwise devoid
of problems, it does appear to do what is asked of it, except in cases
where it hits the end-of-file, where it does get a bit messy, but you
presumably aren't worried about that.

For the record, here's how I'd have written it:

#include <stdio.h>

int main(void)
{
int ch = 0;
while((ch = getchar()) != EOF && ch != '\n')
{
putchar(ch);
}

printf("\nThat's it for today.\n");

return 0;
}

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

[mark_bluemel]

On 1 Jun, 11:27, Richard Heathfield <[e-mail]:

[e-mail] said:

On 1 Jun, 10:25, [e-mail] wrote:

Hi,

I'm a newbie programmer. I can't get work the following code.

"I can't get work" doesn't tell me much about your problem.

Actually, I think it tells you all you need to know.

Meow! Harsh but essentially fair, I guess.

However, he did post a complete, compilable, relatively cleanly
formatted program - give him some points for that...

[mark_bluemel]

On 1 Jun, 11:00, [e-mail] wrote:

Hi, Tak,

Your code is much clearner. But the result on my end is the same.

I wanna make the program so that the program prints all the inputs
until a user input a new like (generally by pressing return).

That's what I thought you wanted - why didn't you say so in your
original posting.

What you are saying, I think, is that you want to read and echo the
characters typed as they are typed, without waiting for a line-full of
input. Am I right? I'll assume so.

Do you know, the C language specification doesn't give you a way that
you can guarantee to do this. Input will frequently be buffered, in
some way. You might spend some time with the FAQ at c-faq.com,
especially http://c-faq.com/stdio/index.html questions 12.1 and 12.5,
to understand about this and other issues with your code.

So, it should open for a new input.

That string "That's it for today" should be displayed only when a user
type a newline.

[Samuel.Coddler]

Thanks, guys. Sorry for my ill-mannered post.

I started programming a few days ago. I will read more about what you
wrote.

I need to spend more time before understanding your posts.

Regards,

[CBFalconer]

Sa************@gmail.com wrote:

>
I'm a newbie programmer. I can't get work the following code.

/*
Objective:

Use a loop to print out all of the input characters until a newline
is found.

*/

#include <stdio.h>

int main (){
/* Declare variable, Assign a initial value x. */
char a = 'x';

while (a != '\n') {
/* Get a user input */
a = getchar();
if (a == '\n'){
/* If the input is a new line, say good bye to user */
printf("\nThat's it for today.\n");

} else {
/* Prints a user input */
printf("%c",a);
}
}
return 1;
}

Here's a slightly different program, which should work. Note the
proper indentation etc.

#include <stdio.h>

int main (void) {
int a;

while ((EOF != (a = getchar()) && (a != '\n'))
printf("%c",a);
printf("\nThat's it for today.\n");
return 0
} /* untested */

--
<http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt>
<http://www.securityfocus.com/columnists/423>
<http://www.aaxnet.com/editor/edit043.html>
<http://kadaitcha.cx/vista/dogsbreakfast/index.html>
cbfalconer at maineline dot net

--
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[CBFalconer]

ma**********@pobox.com wrote:

>

.... snip ...

>
I'd better qualify my comments by saying that I didn't bother
with the nitpicking about whether you should declare main as
"int main(void)" or that a should be an int.

None of that is a nit.

--
<http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt>
<http://www.securityfocus.com/columnists/423>
<http://www.aaxnet.com/editor/edit043.html>
<http://kadaitcha.cx/vista/dogsbreakfast/index.html>
cbfalconer at maineline dot net

--
Posted via a free Usenet account from http://www.teranews.com

[Barry Schwarz]

On Fri, 01 Jun 2007 09:25:00 -0000, Sa************@gmail.com wrote:

>Hi,

I'm a newbie programmer. I can't get work the following code.

snip code

It helps a whole bunch if you tell us what is happening that is
different than what you expect.
Remove del for email